Galileo discovered during his inclined-plane experiments that a ball rolling down an incline and onto a horizontal surface would roll indefinitely.
The simplest way to do this is to set up equivalent fractions, like this-
![\frac{1}{2.2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2.2%7D%20)
=
![\frac{40}{x}](https://tex.z-dn.net/?f=%20%5Cfrac%7B40%7D%7Bx%7D%20)
Solve for x by using cross multiplication.
40*2.2= 88
1*x=88
x=88
Therefore, the boy weighs 88lbs.
Answer:
![T_f=24.71](https://tex.z-dn.net/?f=T_f%3D24.71)
Explanation:
From the question we are told that:
Mass of block ![m=50](https://tex.z-dn.net/?f=m%3D50)
Temperature of block ![T_b =140 \textdegree C](https://tex.z-dn.net/?f=T_b%20%3D140%20%5Ctextdegree%20C)
Volume of water ![V= 90L](https://tex.z-dn.net/?f=V%3D%2090L)
Temperature of water ![T_w=10 \textdegree C](https://tex.z-dn.net/?f=T_w%3D10%20%5Ctextdegree%20C)
Density of water ![\rho=1000kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D1000kg%2Fm%5E3)
Specific heat of water ![C_w=4.18KJ/kg-k](https://tex.z-dn.net/?f=C_w%3D4.18KJ%2Fkg-k)
Specific heat of copper ![C_p=0.96KJ/kg-k](https://tex.z-dn.net/?f=C_p%3D0.96KJ%2Fkg-k)
Generally the equation for equilibrium stage is mathematically given by
![mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)](https://tex.z-dn.net/?f=mC_p%28T_b-T_f%29%3D%5Crho%2AVV%2Ac%28T_f-T_w%29)
![50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)](https://tex.z-dn.net/?f=50%2A0.96%28140-T_f%29%3D1000%2A90%2A10%5E-3%2Ac_w%28T_f-10%29)
![48(140-T_f)=376.2(T_f-10)](https://tex.z-dn.net/?f=48%28140-T_f%29%3D376.2%28T_f-10%29)
![140-T_f=7.8375(T_f-10)](https://tex.z-dn.net/?f=140-T_f%3D7.8375%28T_f-10%29)
![140-T_f=7.8375T_f-78.375](https://tex.z-dn.net/?f=140-T_f%3D7.8375T_f-78.375)
![-8.8375T_f=-218.375](https://tex.z-dn.net/?f=-8.8375T_f%3D-218.375)
![T_f=\frac{-218.375}{-8.8375}](https://tex.z-dn.net/?f=T_f%3D%5Cfrac%7B-218.375%7D%7B-8.8375%7D)
![T_f=\frac{-218.375}{-8.8375}](https://tex.z-dn.net/?f=T_f%3D%5Cfrac%7B-218.375%7D%7B-8.8375%7D)
![T_f=24.71](https://tex.z-dn.net/?f=T_f%3D24.71)
Answer:
it will be 1/√2 of its original period.
Explanation: